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2.3b Problem Solving With Quadratic Functions and Equations:

       Finding Max and Min Values: Worked Examples

Example 1.  Determine the coordinates of the vertex for the graph of .

As  and , we have .

Now  

So the coordinates of the vertex are  

Example 2.  Determine the coordinates of the vertex and intercepts for the graph of .

We have  

Now  

Thus  

The coordinates of the vertex are .

To find the y-intercept, we set  x equal to zero and evaluate:

The y-intercept is (0,25)

To find the x-intercepts, we set the output (the y-coordinate) equal to zero and solve:

The x-intercept is (5,0).

Example 3.  Determine the  vertex and intercepts for the graph of  

We have  

Now  

And  

So the coordinates of the vertex are .

As  

So the coordinates of the y-intercept are .

 Attempting to solve the equation  using the quadratic formula we have:

As there are no real solutions to the above equation, there are no x-intercepts.

Example 4.  The profit earned by a company for manufacturing and selling  units of particular product is measure by the function , where  is in hundreds of dollars. Determine the number of units the company needs to manufacture and sell to maximize its profit and determine the maximum profit.

We have   

So,

Thus, the company should produce and sell 100 units to maximize profit.

Now,    

Hence the maximum profit would be $5,500,000

Example 5. A farmer has 3600 feet of fencing and wishes to enclose a rectangular area. What is the maximum area she can enclose?

We know the area of a rectangle can be measured by multiplying its length by its width.

Let  represent the length of the rectangle  and let  represent its width.  We start with the formula .

We wish to create a function that measures the area of a rectangle. We shall create a function that measures a rectangles area based upon its length. So, we need to rewrite the rectangles width  in terms of its length . To do this we consider the perimeter of the rectangle which can be calculated by summing the sides of the rectangle or equivalently using the formula . We have

Now we can replace  in the area formula with  to create the function , which indeed measures the area of the rectangle based upon its length.

We have

So,  

The length of the rectangle should be 900 feet in order to maximize its area.

Now,  

Thus the maximum area would be 810,000 square feet.

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